Question #143452
Calculate the potential of the following concentration cell at 25.0 ◦C.
Ag | Ag+(1.00×10−3 M) || Ag+(0.100 M) | Ag
Identify the anode and the cathode in the cell. How would the cell potential change if the concentrations
of Ag+ in both half-cells were doubled? Use the Nernst equation to support your answer.
Ag | Ag+(1.00×10−3 M) || Ag+(0.100 M) | Ag
Identify the anode and the cathode in the cell. How would the cell potential change if the concentrations
of Ag+ in both half-cells were doubled? Use the Nernst equation to support your answer.
Expert's answer
Ag | Ag+(1.00×10−3 M) || Ag+(0.100 M) | Ag
concentration cell at 25.0 ◦C.
An electrochemical cell of this type, in which the anode and cathode compartments are identical except for the concentration of a reactant, is called a concentration cell. As the reaction proceeds, the difference between the concentrations of Ag+ in the two compartments will decrease, as will Ecell.
2Ag+(aq) +2e- > 2Ag(s)
Cu(s) > Cu2+(aq) + 2e-
Eo reduction = +0.799v
Eo Oxidation = - 0.518v
2Ag+(aq) + Al(s) > 2Ag(s) + Cu2+(aq) EO =+0.281v
N = 2 moles of electrons
Eocell = (0.0592/n) log K
0.281 = (0.0592/2) log K
Log k = 9.49
K = 109.49 =3.1 x 109
